PROFESSOR: Welcome back to recitation.
In this video segment, what I'd like us to do is work on
this following problem.
Find d dx of the integral from 0 to x squared, cosine t dt.
I'm going to give you a moment to think about it and then
I'll come back and show you how I do it.
OK, welcome back.
Hopefully you were able to make some headway on this.
Let's look at the problem and see how we
would break it down.
Well we know from the fundamental theorem of
calculus that you saw in the lecture, that if up here
instead of x squared we had an x, then the problem would be
easy to solve.
We'd just use the fundamental theorem of calculus, the
answer would be cosine x.
But of course, we don't have an x, we have an x squared.
That's why I gave you this problem.
And we need to figure out how to solve this problem when
there's a different function up here besides just x.
What we're going to use is we're going to combine the
fundamental theorem of calculus and the chain rule.
So, let's start off with how we would do this if it were
the integral from 0 to x, as I mentioned.
So we'll define capital F of x to be equal to the integral
from 0 to x cosine t dt.
And then we know that f prime of x is equal to cosine x.
Now the problem is, we don't just
have this, as I mentioned.
What we actually have--
let me write this down--
we have f of x squared.
Right?
That's what this-- sorry let me highlight what I mean--
this boxed thing is f of x squared.
So we took f of x and we evaluated it at x squared.
That's what we get in the box.
And so if we want to find d dx of f of x squared, it really
is just the chain rule.
We really want to think of this as a
composition of functions.
The first, the outside function is capital F and the
inside function is x squared.
So just in general, how do we think about the chain rule?
Well remember what we do--
let me come back here for a second--
remember what we do is we take the derivative of the outside
function, we evaluate it at the inside function and then
we take the derivative of the inside function and we
multiply those two together.
So all I have to do is figure out, is what is
the following thing?
We know d dx, the quantity f of x squared should be equal
to F prime evaluated at x squared times 2x.
That's just what we said earlier, right?
It's the derivative of F evaluated at x squared times
the derivative of x squared.
So now I just have to figure out what this is.
Well let's go back to the other side and see what we
wrote F prime was.
If we come over here, we see F prime at x is just cosine x.
So F prime at x squared is going to be this function
evaluated at x squared.
That's just cosine of x squared.
So we see over here, we just get cosine x squared times 2x.
And because I'm a mathematician, I want to write
the 2x in front before I finish.
Becaues otherwise I get confused.
So the answer here is just 2x times cosine x squared.
Now I want to point out really what we did here.
This is the answer to this particular problem, but we can
now solve problems in general when I put any function up
here, any function of x up here.
Right?
Ultimately, all I did was I used the fundamental theorem
of calculus and the chain rule.
So any function I put up here, I can do
exactly the same process.
I would define f of x to be this type of thing, the way we
would define it for the
fundamental theorem of calculus.
I would know what F prime of x was.
And then I would have to evaluate F at a, at this
function up here, whatever I put up there.
So in this case it was x squared.
I could have made it natural log x.
I could've made it some big polynomial or something more
complicated.
Right?
And once I do that, I just follow this same process.
Now, instead of the x squared here I would
have that other function.
So I'd evaluate capital F at whatever function that is
times the derivative of that function.
It's exactly the same process.
So I want to point out that this is a bigger situation
than I had before, or a bigger situation than just this
little problem.
So, just so you understand that.
OK?
So again, I just want to say one more time.
Now you know how to solve problems where you have any
other function of x up here and you want to take the
derivative of this kind of expression of an integral with
another function of x up there.
All right, I think I'll stop there.